Compute the number of ordered pairs of positive integers (a,b) that satisfy (a^2)(b^3)=(20)^18.
+8 That question has been solved here: https://web2.0calc.com/questions/algebra-question_67
+2 (a^2)(b^3)=(20)^18.
20^18 =2^36 * 5^18
a =(2^18)^2 =262,144^2
b =(5^6)^3 =15,625^3
20^18==2.62144e+23 = 2^36 * 5^18
Total number of prime factors= 54
Total number of ordered pairs= 2,701
20^18==2.62144e+23 = 2^36 * 5^18
Total number of prime factors= 54
Total number of ordered triplets= 4,104,919
a=262,144
b=15,625
LCM[15,625, 262,144)=11,790,000
11790000 = 2^4 * 3^2 * 5^4 * 131
Number of ordered pairs=[ (4*2+1) + (2*2+1) + (4*2+1) + (2*1+1)]==9 * 5 * 9 *3==1,215
+118609
Compute the number of ordered pairs of positive integers (a,b) that satisfy (a^2)(b^3)=(20)^18.
\( 20^{18}=2^{36}*5^{18}\)
\(5^0*5^{18}\\ 5^6*5^{12}\\ \\~\\ 2^0*2^{36}\\ 2^6*2^{30}\\ 2^{12}*2^{24}\\ 2^{18}*2^{18}\\ \)
5^0 can be paired with 7 of them (I mean different powers of 2)
5^6 can be paired with 7 of them
So I just get 14 but order counts, and a and b are all interchangeable so that is 28
Here they are
If you can spot any that I have missed then please let me know.
(I am not guarenteeing that this answer is right)
